อนุกรมนี้แก้อย่างไงคะ

[natto]

k-1
S3/2 i²(1/2)ⁱ
i=0
ทำอย่างไรจึงได้คำตอบอย่างนี้
-6[(1/2k²+k+3/2)(1/2)^k-3/2]

[Mastermander]

ขอโทษครับมาช้าไป
กำลังทำให้อยู่ครับ

[Mastermander]

$$ \sum_{i=0}^{k-1}\frac{3}{2}i^2\bigg(\frac{1}{2}\bigg)^i =\frac{3}{2} \sum_{i=0}^{k-1}i^2\bigg(\frac{1}{2}\bigg)^i $$

$$ \sum_{i=0}^{k-1}\frac{3}{2}i^2\bigg(\frac{1}{2}\bigg)^i = \frac{3}{2}\Bigg(0+1(\frac{1}{2})+4(\frac{1}{2})^2+\dots+(k-1)^2(\frac{1}{2})^{k-1}\Bigg)$$

$$ให้\quad S_k=\Bigg(1(\frac{1}{2})+4(\frac{1}{2})^2+\dots+(k-1)^2(\frac{1}{2})^{k-1}\Bigg)\quad ...(1)$$

$$\frac{1}{2}S_k = \Bigg(1(\frac{1}{2})^2+4(\frac{1}{2})^3+\dots+(k-1)^2(\frac{1}{2})^k\Bigg) \quad ...(2)$$

\[ (1)-(2);\frac{1}{2}S_k=\Bigg(\frac{1}{2}+3(\frac{1}{2})^2+5(\frac{1}{2})^3+\dots+(2k-3)(\frac{1}{2})^{k-1}-(k-1)^2(\frac{1}{2})^k\Bigg) \]

\[\frac{1}{2}S_k=\Bigg(\bigg(\frac{1}{2}+3(\frac{1}{2})^2+5(\frac{1}{2})^3+\dots+(2k-3)(\frac{1}{2})^{k-1}\bigg)-(k-1)^2(\frac{1}{2})^k\Bigg) \]

$$ให้ \quad S_g=\bigg(\frac{1}{2}+3(\frac{1}{2})^2+5(\frac{1}{2})^3+\dots+(2k-3)(\frac{1}{2})^{k-1}\bigg)\quad ...(3)$$

$$\frac{1}{2}S_g=(\frac{1}{2})^2+3(\frac{1}{2})^3+5(\frac{1}{2})^4+\dots+(2k-3)(\frac{1}{2})^{k} \quad ...(4)$$

$$(3)-(4);\frac{1}{2}S_g=\bigg(\frac{1}{2}+2(\frac{1}{2})^2+2(\frac{1}{2})^3+\dots+2(\frac{1}{2})^{k-1}-(2k-3)(\frac{1}{2})^k\bigg)$$

$$\frac{1}{2}S_g=\bigg(\frac{1}{2}+2\Big((\frac{1}{2})^2+(\frac{1}{2})^3+(\frac{1}{2})^4+\dots+(\frac{1}{2})^{k-1}\Big)-(2k-3)(\frac{1}{2})^k\bigg)$$

$$\frac{1}{2}S_g=\Bigg(\frac{1}{2}+2\bigg(\frac{\frac{1}{4}\big(1-(\frac{1}{2})^{k-2}\big)}{1-1/2}\bigg)-(2k-3)(\frac{1}{2})^k\Bigg)$$

$$\frac{1}{2}S_g=\bigg(\frac{1}{2}+1-4(\frac{1}{2})^k-(2k-3)(\frac{1}{2})^k\bigg)$$

$$S_g=\bigg(3-(4k+2)(\frac{1}{2})^k\bigg) $$

$$S_k=2\Bigg(\bigg(3-(4k+2)(\frac{1}{2})^k\bigg)-(k-1)^2(\frac{1}{2})^k\Bigg)$$

$$S_k=2\Big(3-(k^2+2k+3)(\frac{1}{2})^k\Big)$$

$$\sum_{i=0}^{k-1}\frac{3}{2}i^2\bigg(\frac{1}{2}\bigg)^i = \frac{3}{2}S_k$$

$$\sum_{i=0}^{k-1}\frac{3}{2}i^2\bigg(\frac{1}{2}\bigg)^i = -3\bigg((k^2+2k+3)(\frac{1}{2})^k-3\bigg)$$